4
$\begingroup$

Let $W(t_1,\dotsc,t_n)$ a holomorphic function on some connected open set $U$ of $\mathbb C^n$. Let $\mathbf t^{(0)}$ a point of $U$.

Assume that there exists a cycle $\gamma$ in $\mathbb C^m$ and a rational function $F(\mathbf t, x_1,\dotsc, x_m)$ such that for all $\mathbf t$ in a neighbourhood of $\mathbf t^{(0)}$:

  1. $\mathbf x \mapsto F(\mathbf t, \mathbf x)$ is continuous on $\gamma$;
  2. $\displaystyle W(\mathbf t) = \oint_\gamma F(\mathbf t, \mathbf x)\mathrm d \mathbf x$.

Is it true that for all point $\mathbf t^{(1)}$ in $U$, there exists another (sum of) cycle $\gamma_1$ such that properties 1. and 2. are satisfies in a neighbourhood of $\mathbf t^{(1)}$ ?


For simple integrals, that is $m=1$, this is true, and not too hard to see: the poles of $F(\mathbf t, x)$ are points that move continuously with $\mathbf t$. It is easy to deform $\gamma$ so that is does not encounter these moving points. There is a singularity when a pole inside $\gamma$ collapses with a pole outside, but this can't happen if we stay in the domain of holomorphy of $W$. However, it seems harder for multiple integrals...

$\endgroup$
1
3
+200
$\begingroup$

I'm not sure you're right about the $m=1$ case. Take $$F(t,x) = \frac{2x^3}{x^2-t} = \frac{x^2}{x-\sqrt{t}} + \frac{x^2}{x+\sqrt{t}}$$ take $t^0=1$ and take $\gamma$ to be a circle in the $x$-plane which encloses $1$ but not $-1$. Then $$\oint_{\gamma} F(x,t) dx = (2 \pi i) (\sqrt{t})^2 = (2 \pi i) t$$ where $\sqrt{t}$ means the square root of $t$ which lies in $\gamma$.

Now, the function $W(t) = (2 \pi i) t$ extends holomorphically to the whole $t$-plane. Consider $t^1$ to be the point $t=0$.

If I understand correctly, you are claiming that there is a cycle $\gamma^1$ so that $F(t,x)$ will be continuous on $t \times \gamma^1$ for any $t$ near $0$, and this cycle will have $\int_{\gamma_1} F(t,x) dx = (2 \pi i) t$.

But $F$ is discontinuous at $(0,0)$, so $\gamma_1$ must not pass through $0$. Thus, either $\gamma_1$ encloses $0$ or it doesn't. So, for $t$ near but not equal to $0$, either $\gamma_1$ encloses both poles of $F$ or neither one, and we get $\oint_{\gamma_1} F = 2 (2 \pi i) t$ or $0$.

In other words, I disagree with your statement "There is a singularity when a pole inside $\gamma$ collapses with a pole outside, but this can't happen if we stay in the domain of holomorphy of $W$." If I rig the pole residues carefully, I can make $W$ stay holomorphic even though the pieces it is built out of are becoming singular.

$\endgroup$
3
  • $\begingroup$ In retrospect, an easier example is $F(t,x) = (2x)/(x^2-t) = 1/(x-\sqrt{t}) + 1/(x+\sqrt{t})$. $\endgroup$ Jan 27 '14 at 18:17
  • $\begingroup$ Still seems impossible, with the same example. How do you get a contour $\gamma$ which, for $t$ in a punctured neighborhood of $0$, always has precisely one of $\pm \sqrt{t}$ inside $\gamma$, with $\pm \sqrt{t}$ never landing on $\gamma$ itself? $\endgroup$ Jan 27 '14 at 22:22
  • $\begingroup$ Damn! You got this one ;) Too bad the answer is not positive. However, if you allow linear combination of cycle, then you can choose $\gamma_1$ to be $\frac 12$ times the unit circle. But I'm not sure this is a real objection. (And you right for the punctured disc, that was pointless ;) $\endgroup$
    – Lierre
    Jan 27 '14 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.